EITG05 Digital Communications

Fourier transform EITG05 Digital Communications Lecture 4 Bandwidth of Transmitted Signals Michael Lentmaier Thursday, September 3, 08 X(f )F{x(t)} x(t) e jπ ft dt X Re (f )+jx Im (f ) X(f ) e jϕ(f ) x(t)f {X(f )} X(f ) e +jπ ft df X(f ) e +j(π ft+ϕ(f )) df Some useful Fourier transform properties g(at) a G(f/a) g( t) G( f) G(t) g( f) g(t t 0 ) G(f)e jπft0 g(t)e jπfct G(f f c ) d g(t) dt jπf G(f) g (t) G ( f) g (T t) G (f)e jπft δ(t) full list in Appendix C of the compendium (dc) δ(f) e jπfct δ(f f c ) cos(πf c t) (δ(f + f c)+δ(f f c )) sin(πf c t) j (δ(f + f c) δ(f f c )) αe πα t e πf /α Some useful Fourier transform properties Consider two signals x(t) and y(t) and their Fourier transforms x(t) X(f ), y(t) Y(f ) Recall the convolution operation z(t) x(t) y(t): x(t) A t τ z(t) AB τ τ y(t) B z(t) x(t) y(t) + t 0τ τ t 0τ y(ν) x(t ν) dν Filtering: x(t) y(t) X(f ) Y(f ) Multiplication: x(t) y(t) X(f ) Y(f )

Spectrum of time-limited signals Consider some time-limited signal s T (t) of duration T, with s T (t)0 for t < 0 and t > T Assume that within the interval 0 t T, the signal s T (t) is equal to some signal s(t), i.e., s T (t)s(t) g rec (t), where g rec (t) is the rectangular pulse of amplitude A Taking the Fourier transform on both sides we get S T (f )S(f ) G rec (f )S(f ) AT sin(π ft) π ft jπ ft e Since G rec (f ) is unlimited along the frequency axis, this is the case for S T (f ) as well (convolution increases length) Time-limited signals can never be strictly band-limited Some definitions of bandwidth Main-lobe definition: W lobe is defined by the width of the main-lobe of R(f ) This is how we have defined bandwidth in previous examples In baseband we use the one-sided width, while in bandpass applications the two-sided width is used (positive frequencies) Percentage definition: W 99 is defined according to the location of 99% of the power For bandpass signals W 99 is found as the value that satisfies fc +W 99 / f c W 99 / R(f )df 0.99 0 R(f )df Other percentages can be used as well: W 90, W 99.9 Nyquist bandwidth Assuming an ideal pulse with finite bandwidth (see Chapter 6) W nyq R s [Hz] Some definitions of bandwidth Pulse spectrum examples Pulse shape W lobe %power W 90 W 99 W 99.9 Asymptotic in W lobe decay rec /T 90.3.70/T 0.6/T 04/T f tri 4/T 99.7.70/T.60/T 6.4/T f 4 hcs 3/T 99.5.56/T.36/T 5.48/T f 4 rc 4/T 99.95.90/T.8/T 3.46/T f 6 Nyquist R s 00 0.9R s 0.99R s 0.999R s ideal Table.: Double-sided bandwidth results for power spectral densities according to (.). The g rec (t), g tri (t), g hcs (t) andg rc (t) pulse shapes are defined in Appendix D, and T denotes the duration of the pulse. The Nyquist pulse shape is not limited in time and it is defined in (D.49) with parameters β 0and T. G(f) EgT in db: rec, hcs, and rc pulse 0 0 0 30 40 50 60 70 ft 3 4 5 6 7 8 This table is useful for PAM, PSK, and QAM constellations Except bandwidth W, the asymptotic decay is also relevant ( ) Figure.9: 0 log G(f) 0 E gt for the g rec (t), g hcs (t), and g rc (t) pulse shapes. See also Example.6.

From last lecture: R(f ) for Binary Signaling In the general binary case, i.e., M, wehave A(f )P 0 S 0 (f )+P S (f ) This simplifies the expression for the power spectral density to R(f )R c (f ) + R d (f ) P 0P S 0 (f ) S (f ) + P 0 S 0 (f )+P S (f ) T b Tb δ(f n/t b ) n (derivation in Ex..0) We will now consider some examples from the compendium Example. Assume equally likely antipodal signal alternatives, such that s (t) s 0(t) g(t) where g(t) g rec(t), and g rec(t) is given in (D.). Assume also that T T b. i) Calculate the power spectral density R(f). ii) Calculate the bandwidth W defined as the one-sided width of the mainlobe of R(f ), if the information bit rate is 0 [kbps], and if T T b/. Calculate also the bandwidth efficiency ρ. iii) Estimate the attenuation in db of the first sidelobe of R(f) compared to R(0). M with equally likely antipodal signaling s (t) s 0 (t)g(t) With P 0 P / and S (f ) S 0 (f )G(f ) we get R(f )R b S (f ) R b S 0 (f ) R b G(f ) Details for the pulse in Appendix D Example.3 Assume equally likely antipodal signal alternatives below. Assume that s (t) s 0(t) g rc(t), where the time raised cosine pulse g rc(t) is defined in (D.8). Assume also that T T b. Find an expression for the power spectral density R(f). Calculate the bandwidth W, defined as the one-sided width of the mainlobe of R(f), ifr b is 0 [kbps]. Calculate also the bandwidth efficiency ρ. Same as Example., but with g rc (t) pulse Analogously we get R(f )R b G rc (f ) From the one-sided main-lobe we get W /T [Hz] Bandwidth efficiency ρ / [bps/hz] is the same (why?) Example.4 Assume P 0 P and that, s (t) s 0(t) g rc(t)cos(πf ct) with T T b, and f c /T. Hence, a version of binary PSK signaling is considered here (alternatively binary antipodal bandpass PAM). Calculate the bandwidth W, defined as the double-sided width of the mainlobe around the carrier frequency f c. Assume that the information bit rate is 0 [kbps]. Calculate also the bandwidth This corresponds to the bandpass case Let g hf (t) denote the high-frequency pulse g hf (t)g rc (t)cos(π f c t) and R(f )R b G hf (f ) Using shift operations we get G rc (f + f c ) R(f )R b + G rc(f f c ) From the two-sided main-lobe we get W 4/T [Hz]

Example: discrete frequencies in R(f ) Assume M Let s 0 (t)0 and s (t)5 with a pulse duration T T b / With this the average signal becomes a(t) s 0(t)+s (t).5, 0 t T We can then write (within the pulse duration T) s 0 (t).5 + a(t), s (t)+.5 + a(t) Observe:. this method is a waste of signal energy since a(t) does not carry any information. repetition of a(t) in every symbol interval creates some periodic signal component in the time domain, which leads to discrete frequencies in the frequency domain From last lecture: general R(f ) The power spectral density R(f ) can be divided into a continuous part R c (f ) and a discrete part R d (f ) R(f )R c (f )+R d (f ) The general expression for the continuous part is R c (f ) M P n S n (f ) A(f ) ( n0 For the discrete part we have R d (f ) ) M P n S n (f ) A(f ) n0 A(f ) T s δ(f n/ ) n R(f ): M-ary PAM signals With M-ary PAM signaling we have s l A l g(t), l 0,,...,M Then M S l (f )A l G(f ), and A(f ) l0 With this we obtain the simplified expression R(f ) σ A G(f ) + m A T s G(f ) n P l A l G(f ) δ(f n/ ), where m A denotes the mean and σa E s/e g m A the variance of the amplitudes A l Assuming zero average amplitude m A 0 and using P σa E g R s this reduces to R(f )R c (f ) σ A G(f ) P E g G(f ) Example.8 Assume the bit rate R b 9600 [bps], M-ary PAM transmission and that m A 0. Determine the (baseband) bandwidth W, defined as the one-sided width of the mainlobe of the power spectral density R(f), ifm, M 4and M 8, respectively. Furthermore, assume a rectangular pulse shape with amplitude A g, and duration T. Calculate also the bandwidth efficiency ρ. What is W for a given pulse shape and different M? Using T, m A 0 and g(t)g rec (t), wehave R(f ) σ A G rec (f ) For the given pulse we get W /, where kt b k M W 9600[Hz] k M 4 W 4800[Hz] k 3 M 8 W 300[Hz] Bandwidth efficiency: ρ R b /W kt b /T b k

What does bandwidth efficiency tell us? In the previous example we had a bandwidth efficiency of ρ R b W k Saving bandwidth The previous example showed that the bandwidth W can be reduced by increasing M T kt b increases with M W /T R b /k decreases accordingly Improving bit rate Assume instead that the bandwidth W is fixed in the same example, i.e., the symbol duration T is fixed Then R b kw increases with M Assume for example W MHz: R b Mbps if M (k ) R b 0Mbps if M 04 (k 0) R(f ): M-ary QAM signals With M-ary QAM signaling the signal alternatives are s l (t)a l g(t) cos(π f c t) B l g(t) sin(π f c t), l 0,,...,M Then the Fourier transform becomes G(f + f c )+G(f f c ) G(f + f c ) G(f f c ) S l (f )A l jb l (A l jb l ) G(f + f c) +(A l + jb l ) G(f f c) Assuming a zero average signal a(t)0 and f c T this simplifies to R(f )R c (f )P G(f + f c) + G(f f c ) E g R(f ): M-ary QAM signals Remember that M-ary QAM signals contain M-ary PSK and M-ary bandpass PAM signals as special cases: BP-PAM: B l 0 PSK: A l cos(ν l ), B l sin(ν l ) our results for R(f ) of M-ary QAM signals include these cases For symmetric constellations, such that a(t)0, the simplified version applies The bandwidth W is determined by G(f f c ) and hence the two-sided main-lobe of G(f ) if the same pulse g(t) is used then M-ary QAM, M-ary bandpass PAM and M-ary PSK have the same bandwidth W Example Bandwidth consumption for BPSK, QPSK and 6-QAM assuming equal R b and f c 00R b RbR(f)/ P in db: M-ary QAM and ghcs(t) 0 0 0 30 40 50 f/r b 99 00 0 0 Figure.0: The power spectral density for binary QAM (BPSK, widest mainlobe), 4-ary QAM (QPSK), and 6-ary QAM (smallest mainlobe). The figure shows 0 log 0 (R b R(f)/ P ) [db] in the frequency interval 98R b f 0R b. The carrier frequency is f c 00R b [Hz], and a kt b long g hcs (t) pulse is assumed. See also (.7) and (.30).

R(f ): M-ary FSK signals With M-ary frequency shift keying (FSK) signaling the signal alternatives are s l (t)a cos(π f l t + ν), Choosing ν π/ this can be written as 0 t s l (t)g rec (t) sin(π f l t), with T, since s l (t)0 outside the symbol interval The Fourier transform is then S l (f )j G rec(f + f l ) G rec (f f l ) The exact power spectral density R(f ) can now be computed by the general formula (.0) (.04) R(f ): M-ary FSK signals Let us find an approximate expression for the FSK bandwidth W Assume that f l f 0 + l f, l 0,...,M Then the bandwidth W can be approximated by W R s + f M f 0 + R s (M )f + R s Consider now orthogonal FSK with f I R s / for some I > 0 The bandwidth efficiency is then ρ R b W R b (M )f + R s R b ( (M )I/ + ) Rs log M (M )I/ + Observe: the bandwidth efficiency of orthogonal M-ary FSK gets small if M is large Last week we saw: M-ary FSK has good energy and Euclidean distance properties trade-off Example.36 Assume that orthogonal M-ary FSK is used to communicate digital information in the frequency band. f. [MHz]. For each M below, find the largest bit rate that can be used (use bandwidth approximations): i) M ii) M 4 iii) M 8 iv) M 6 v) M 3 Which of the M-values above give a higher bit rate than the M case? Solution: It is given that W M FSK 00 [khz]. From (.45), the largest bit rate is obtained with I : R b 0 5 log (M) (M )/ + R(f ): OFDM-type signals An OFDM symbol (signal alternative) x(t) can be modeled as a superposition of N orthogonal QAM signals, each carrying k n bits, that are transmitted at different frequencies (sub-carriers) N x(t) s n,qam (t) n0 Assuming each QAM signal has zero mean and that the different carriers have independent bit streams we get M 4 8 6 3 log (M) (M )/+ 5/ R b 0.4 40 kbps 4 0.574 57 kbps 7/ 7 3 6 0.5455 55 kbps / 4 8 0.4 4 kbps 9/ 9 5 0 0.857 9 kbps 35/ 35 From this table it is seen that M 4, 8, 6 give a higher bit rate than M. R(f )R c (f )R s E{ X(f ) N } R n (f ) n0 Using our previous results for QAM in each sub-carrier we get N R(f )R c (f ) n0 P G(f + f c) + G(f f c ) E g

R(f ): OFDM-type signals Illustration of R n (f ) contributed by three neighboring sub-carriers: Subchannel contributions 0.04 0.03 0.0 0.0 0 96 98 00 0 04 06 ft Assuming f n f 0 + n/( h ) we can estimate the bandwidth as W (N + )f N + R s N R s, N, h h / The bandwidth efficiency is then approximated by ρ R b W R N s W k n k0 N N k0 k n [bps/hz] Example: R(f ) for OFDM R(f)/ PT in db: OFDM with N 6 ft 90 00 0 0 30 0 0 0 30 40 50 N 6 sub-carriers T 0. [ms] f R s /0.95 0.53 [khz] W 0.95 7 R s 79 [khz] R(f)/ PT for OFDM with N 6 0.04 0.03 0.0 0.0 0 90 00 0 0 30 ft Example.35 ADSL: uses plain telephone cable (twisted pair, copper) Power spectral density POTS: telephony, modem, FAX ADSL uplink 64-04 kbps ADSL downlink 0.5-8 Mbps f [khz] 0 4 5 38 04 In ADSL, a coded OFDM technique is used. The level of the power spectral density in the downstream is roughly -73 db. As a basic example, let us here assume that the OFDM symbol rate in the downlink is 4000 [symbol/s], and that the subchannel carrier spacing is 5 khz. Furthermore, it is here also assumed that uncoded 6-ary QAM is used in each subchannel (assumes a very good communication link). For the ADSL downlink above, determine the bit rate in each subchannel, the total bit rate, and the bandwidth efficiency. What about filtering away the side-lobes? Let us use a spectral rectangular pulse X srec (f ) of amplitude A and width f to strictly limit the bandwidth Similar to the time-limited case we can write S f (f )S(f ) X srec (f ) Taking the inverse Fourier transform on both sides we get s f (t)s(t) x srec (t)s(t) Af 0 sin(π f 0 t) π f 0 t Since x srec (t) is unlimited along the time axis, this is the case for the filtered signal s f (t) as well The signal x srec (t) defines the ideal Nyquist pulse As a consequence of filtering, the transmitted symbols will overlap in time domain inter-symbol-interference (ISI)

Nyquist Pulse How can we further improve ρ? a) x 0 /Rnyq R nyq/ X (f) nc R nyq/ f [Hz] b) x (t) nc -T R s x R Ts nyq 0 nyq R nyq Figure 6.6: a) Ideal Nyquist spectrum; b) Ideal Nyquist pulse. t d d n d N t...... α k, α k,n α k,n t MIMO MODEL w w k w N r r...... r k r Nr ML D E C I S I O N R U L E ^ d ^ d... ^ d N t sin(πr nyqt) x nc(t) x 0, t (6.39) πr nyqt { x0/r X nc(f) nyq, f R nyq/ (6.40) 0, f >R nyq/ The Nyquist pulse and the effect of ISI will be studied in Chapter 6 Nt r k α k,nd n + w k n MIMO: multiple-input multiple output transmission over multiple antennas in the same frequency band challenge: the individual wireless channels interfere 5G world record 06: (team from Lund involved) spectral efficiency of 45.6 bps/hz with 8 antennas